Question 18.1: The qualifying examination for a very exclusive preschool in...
The qualifying examination for a very exclusive preschool in the future to which you want to send your child requires that your child answer questions on the kinetic theory of gases. For example, every five-year-old child must (Figure 18.2) be able to compute
a. The root mean square molecular velocity.
b. The total translational internal energy.
for 1.00 kg of nitrogen gas at 20.0°C. How is this done?
a. The kinetic theory root mean square molecular velocity is given by Eq. (18.13) as
V_\text{rms} = \sqrt{3RT}
For nitrogen, R = 296 \text{J/kg.K} = 296 \text{N.m/kg.K} = 296 \text{m}^2 /\text{s}^2.\text{K} (from Table C.15b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics), then Eq. (18.13) gives
V_\text{rms} = \sqrt{3[296 \text{m}^2/(\text{s}^2 .\text{K})] (20.0 + 273.15 \text{K})} = 510. \text{m/s}
b. The kinetic theory total translational internal energy of a gas is given by Eq. (18.14) as
U_\text{trans} = \frac{3}{2} NkT
Since the total mass of gas present is m_\text{total} = m_\text{molecule} N, where N is the number of molecules present, and m_\text{molecule} = molecular mass/Avogadro’s number = M/N_o; for nitrogen,
m_\text{molecule} = \frac{M}{N_o} = \frac{28.0 \text{kg/kgmole}}{6.022 × 10^{26} \text{molecules/kgmole}} = 4.65 × 10^{−26} \text{kg/molecule}
then,
N = \frac{m}{m_\text{molecule}} = \frac{1.00 \text{kg}}{4.65 × 10^{−26} \text{kg/molecule}} = 2.15 × 10^{25} \text{molecules}
The total translational internal energy in 1.00 \text{kg} of nitrogen gas is
U_\text{trans} = \frac{3}{2} NkT = \frac{3}{2} [(2.15 × 10^{25} \text{molecules}) (1.380 × 10^{−23} \frac{\text{J}}{\text{molecule.K}}) (20.0 + 273.15 \text{K})]
= 131,000 \text{J} = 131 \text{kJ}
Note that the kinetic theory internal energy is independent of gas pressure and depends only on gas temperature, which is a characteristic of ideal gas behavior.